For an infinitesimal element we may take $\rho$~constant and insert limits of integration; but these limits must be taken for the time~$t - r$, and this introduces an important factor representing a kind of Doppler effect. If the element of charge is bounded by two planes perpendicular to the direction of~$r$, the limits of integration are from the front plane at time $t - r$ to the rear plane \PageSep{179} \index{Retarded potential}% at time $t - r - dr$. If $v_{r}$~is the component velocity in the direction of~$r$, the front plane has had time to advance a distance~$v_{r}\, dr$. Consequently the instantaneous thickness of the element of charge is less than the distance between the limits of integration in the ratio $1 - v_{r}$; and the integration is over a volume $(1 - v_{r})^{-1}$~times the instantaneous volume of the element of charge. Hence \[ \iiint \rho\, d\xi\, d\eta\, d\zeta = \frac{de}{1 - v_{r}}. \] Writing as usual $\beta$~for the FitzGerald factor~$dt/ds$, \Eq{(74.62)}~becomes \[ \kappa^{\mu} = \left\{\frac{A^{\mu}\, de}{4\pi r\beta(1 - v_{r})}\right\}_{t-r} = \left\{\frac{de(u, v, w, 1)}{4\pi r(1 - v_{r})}\right\}_{t-r}. \Tag{(74.71)} \] In most applications the motion of the charge can be regarded as uniform during the time of propagation of the potential through the distance~$r$. In that case \[ \{r(1 - v_{r})\}_{t-r} = \{r\}_{t}, \] the present distance being less than the antedated distance by~$v_{r} r$. The result then becomes \[ \kappa^{\mu} = \left\{\frac{A^{\mu}\, de}{4\pi r\beta}\right\}_{t} = \left\{\frac{de(u, v, w, 1)}{4\pi r}\right\}_{t}. \Tag{(74.72)} \] It will be seen that the scalar potential~$\Phi$ of a charge is unaltered by uniform motion, and must be reckoned for the present position of the charge, \emph{not from the antedated position}. The equation~\Eq{(74.71)} can be written in the pseudo-tensor form \[ \kappa^{\mu} = \left\{\frac{A^{\mu}\, de}{4\pi A^{\nu}R_{\nu}}\right\}_{R^{\alpha}R_{\alpha}=0}, \Tag{(74.8)} \] where $R^{\mu}$~is the pseudo-vector representing the displacement from the charge \index{Pseudo-vector}% $(\xi, \eta, \zeta, \tau)$ to the point $(x, y, z, t)$ where $\kappa^{\mu}$~is reckoned. The condition $R^{\alpha}R_{\alpha} = 0$ gives \[ -(x - \xi)^{2} - (y - \eta)^{2} - (z - \zeta)^{2} + (t - \tau)^{2} = 0, \] so that \[ \tau = t - r. \] Also \begin{align*} A^{\nu} R_{\nu} &= -\beta u(x - \xi) - \beta v(y - \eta) - \beta w(z - \zeta) + \beta(t - \tau) \\ &= -\beta v_{r} r + \beta r \\ &= r\beta(1 - v_{r}). \end{align*} A \emph{finite} displacement~$R^{\mu}$ is not a vector in the general theory. We call it a pseudo-vector because it behaves as a vector for Galilean coordinates and Lorentz transformations. Thus the equation~\Eq{(74.8)} does not admit of application to coordinates other than Galilean.